# Math Help - help with maximum curvature

1. ## help with maximum curvature

What is the maximal curvature of the curve y = ln cos x?

2. Find an expression for the curvature using

Then find the maximum.

3. ## I'm still having

problems.
y' = -sinx/cosx or -tanx
y'' = -secx^2

4. Put those in to get

$\kappa=\frac{\sec^2 x}{(1+\tan^2 x)^{3/2}}$

Use $1+\tan^2 x=\sec^2 x$

and

$\kappa=\frac{\sec^2 x}{\sec^3 x}$.

So

$\kappa=\frac{1}{\sec x }=\cos x$.